A102357 -id:A102357 - cp's OEIS frontend

A102823 "True already", base 10, start 0: a(n) is the least integer such that the sequence up to a(n-1) written in base 10 contains floor(a(n)/10) copies of the digit a(n) % 10, with a(0) = 0.

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 14, 15, 16, 17, 18, 19, 20, 23, 24, 25, 26, 27, 28, 29, 30, 34, 35, 36, 37, 38, 39, 40, 45, 46, 47, 48, 49, 50, 56, 57, 58, 59, 60, 67, 68, 69, 70, 78, 79, 80, 89, 90, 100, 102, 103, 104, 105, 106, 107, 108, 109, 112, 113

Offset: 0

Hugo van der Sanden, Feb 26 2005

a(n) = A249626(n) for n <= 55. - Reinhard Zumkeller, Nov 03 2014

			The first 11 values of the sequence written in decimal include 2 '1's and 1 '2', so the next value cannot be 11 (the count of '1's is not 1) but can be 12.

Inspired by discussion of "True so far" from Eric Angelini (A102357).

Cf. A102824-A102830, A102357.

Cf. A249626.

A102830 "True already", base 4, start 1: a(n) is the least integer such that the sequence up to a(n-1) written in base 4 contains floor(a(n)/4) copies of the digit a(n) % 4, with a(0) = 1.

Original entry on oeis.org

1, 2, 3, 5, 6, 7, 10, 11, 15, 21, 22, 23, 26, 27, 31, 38, 39, 43, 47, 55, 62, 66, 70, 71, 74, 75, 79, 86, 87, 90, 91, 95, 102, 103, 107, 111, 119, 126, 130, 135, 139, 143, 151, 155, 159, 167, 171, 175, 183, 191, 203, 210, 214, 218, 223, 226, 234, 237, 241, 245, 250

Offset: 0

Hugo van der Sanden, Feb 26 2005

Inspired by discussion of "True so far" from Eric Angelini (A102357).

Cf. A102823-A102829, A102357.

A102850 Non-monotonic "True so far" sequence: In the first n terms, the digit (a(n) mod 10) occurs floor(a(n)/10) times; a(n) is the smallest such number.

Original entry on oeis.org

10, 12, 13, 14, 15, 16, 17, 18, 19, 20, 23, 24, 25, 26, 27, 28, 29, 30, 34, 35, 36, 37, 38, 39, 40, 45, 46, 47, 48, 49, 50, 56, 57, 58, 59, 60, 67, 68, 69, 70, 78, 79, 80, 89, 90, 102, 103, 104, 105, 106, 107, 108, 109, 112, 113, 114, 115, 116, 117, 118, 119, 123, 124

Offset: 1

David Wasserman, Feb 28 2005

Sequence has 5191475 terms. The numbers of occurrences of digits 0-9 are 3589309, 4812817, 4977431, 4564762, 3741602, 3738734, 3599425, 3599878, 3598956, 3589537.
This sequence first differs from the original "True so far" sequence A102357 at a(351) = 920 because this is the first term that is less than the previous term, 1002.
The sequence is injective (no term appears twice) as consequence of the definition, while this is imposed through monotonicity in A102357. - M. F. Hasler, Nov 18 2019

			a(10) = 20 because up to this point there are two 0 digits in the sequence, including the 0 in 20.
a(5191476) doesn't exist. 35893100 would yield a total of 3589311 0's, while 35893110 or 35893120 would yield 3589310 0's. Similar reasons prevent other terms ending with other digits.

M. F. Hasler, Table of n, a(n) for n = 1..10000, Nov 18 2019

Cf. A102357.

c=Vec(0,10); for(n=1,351, a=vecmin(c)*10+10; while(a\10<=c[a%10+1] || a\10 != c[a%10+1]+#select(d->d==a%10,digits(a)), a++); [c[d+1]++|d<-digits(a)]; print1(a",")) \\ M. F. Hasler, Nov 18 2019

A102824 "True already", base 2, start 0: a(n) is the least integer such that the sequence up to a(n-1) written in base 2 contains floor(a(n)/2) copies of the digit a(n) % 2, with a(0) = 0.

Original entry on oeis.org

0, 1, 2, 4, 7, 8, 14, 16, 23, 26, 30, 32, 42, 48, 56, 62, 64, 75, 82, 89, 96, 101, 109, 116, 122, 126, 128, 142, 150, 158, 164, 174, 180, 188, 194, 204, 212, 220, 226, 234, 240, 248, 254, 256, 272, 286, 294, 304, 316, 324, 336, 345, 355, 364, 372, 380, 386

Offset: 0

Hugo van der Sanden, Feb 26 2005

Inspired by discussion of "True so far" from Eric Angelini (A102357).

Cf. A102823-A102830, A102357.

A102827 "True already", base 10, start 1: a(n) is the least integer such that the sequence up to a(n-1) written in base 10 contains floor(a(n)/10) copies of the digit a(n) % 10, with a(0) = 1.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, 22, 23, 24, 25, 26, 27, 28, 29, 33, 34, 35, 36, 37, 38, 39, 44, 45, 46, 47, 48, 49, 55, 56, 57, 58, 59, 66, 67, 68, 69, 77, 78, 79, 88, 89, 99, 111, 112, 113, 114, 115, 116, 117, 118, 119, 122, 123, 124, 125, 126, 127, 128, 129, 133

Offset: 0

Hugo van der Sanden, Feb 26 2005

Conjecture: this sequence in various bases never includes a term divisible by the base.

			The first 9 values of the sequence written in decimal include no '0's and 1 '1', so the next value cannot be 10 (the count of '0's is not 1) but can be 11.

Inspired by discussion of "True so far" from Eric Angelini (A102357).

Cf. A102823-A102830, A102357.

A102827aux := proc(n,dig)
    local c,d ;
    c := 0 ;
    for d in convert(n,base,10) do
        if d = dig then
            c := c+1 ;
        end if;
    end do:
    c ;
end proc:
A102827 := proc(n)
    option remember;
    local a,a10,ad,cum;
    if n < 8 then
        return n+1 ;
    end if;
    for a from 1 do
        a10 := floor(a/10) ;
        ad := a mod 10 ;
        cum := add( A102827aux(procname(i),ad),i=0..n-1) ;
        if cum = a10 then
            return a;
        end if;
    end do:
end proc: # R. J. Mathar, Mar 30 2014

A102829 "True already", base 3, start 1: a(n) is the least integer such that the sequence up to a(n-1) written in base 3 contains floor(a(n)/3) copies of the digit a(n) % 3, with a(0) = 1.

Original entry on oeis.org

1, 2, 4, 5, 8, 13, 14, 17, 23, 29, 32, 35, 41, 44, 50, 56, 62, 67, 74, 76, 82, 88, 92, 95, 98, 104, 110, 113, 116, 122, 125, 131, 137, 143, 152, 161, 173, 179, 188, 193, 202, 206, 215, 223, 226, 232, 238, 244, 250, 256, 263, 269, 274, 278, 284, 287, 293, 299

Offset: 0

Hugo van der Sanden, Feb 26 2005

Inspired by discussion of "True so far" from Eric Angelini (A102357).

Cf. A102823-A102830, A102357.

A102815 "False so far" sequence.

Original entry on oeis.org

11, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 101

Offset: 1

Eric Angelini, Feb 26 2005

The last digit of a(n) is regarded as a glyph and the preceding digits as a number. So "11" reads [one "1"] and "20" [two "0"] - which are both false statements here: there is not one "1" glyph so far in the sequence when [11] is read, but two; and there are not two "0" glyph when [20] is read, but only one. The sequence is built with [a(n+1)-a(n)] minimal and a(n+1) always "false so far". This explains why the sequence doesn't begin with [10]: its statement would be true.
After integer [20] the sequence matches A000027 without [100] -- because [100] would be "true so far".
[This seems to imply that the sequence contains all numbers >= 20 except 100. - N. J. A. Sloane, Aug 22 2011]
The above is correct, and a(n) = n + 19 for n > 81. Proof: It is enough to show that all terms 101 to N are in the sequence, that N >= 101 + 100, and that the count of digits through N is at least floor(N/10) + 11 for each decimal digit. Note that, from the last two digits, the count of each digit increases by at least 20 every 100 numbers, but the required number increases by only 10. Then by a count of the digits, N = 210 suffices. It is easy to check that the remaining conditions are satisfied. - Charles R Greathouse IV, Dec 19 2022

Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A102357.

a(n)=n+if(n>81,19,n>1,18,10) \\ Charles R Greathouse IV, Dec 19 2022

a(n) = n + 19 for n > 81, see comments. - Charles R Greathouse IV, Dec 19 2022

A102825 "True already", base 3, start 0: a(n) is the least integer such that the sequence up to a(n-1) written in base 3 contains floor(a(n)/3) copies of the digit a(n) % 3, with a(0) = 0.

Original entry on oeis.org

0, 1, 2, 3, 5, 6, 9, 11, 14, 17, 18, 24, 25, 27, 31, 38, 41, 42, 45, 50, 51, 54, 63, 68, 69, 72, 78, 81, 88, 94, 98, 104, 108, 110, 113, 116, 122, 125, 129, 132, 135, 143, 147, 150, 153, 159, 162, 174, 180, 188, 192, 198, 204, 207, 213, 216, 225, 231

Offset: 0

Hugo van der Sanden, Feb 26 2005

Inspired by discussion of "True so far" from Eric Angelini (A102357).

Cf. A102823-A102830, A102357.

A102826 "True already", base 4, start 0: a(n) is the least integer such that the sequence up to a(n-1) written in base 4 contains floor(a(n)/4) copies of the digit a(n) % 4, with a(0) = 0.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 7, 8, 11, 12, 16, 18, 19, 22, 23, 26, 27, 31, 32, 39, 40, 43, 44, 48, 55, 56, 60, 64, 69, 70, 74, 75, 79, 86, 87, 90, 91, 95, 96, 103, 104, 107, 108, 112, 119, 120, 124, 128, 138, 139, 143, 151, 152, 155, 156, 160, 163, 167, 171, 172, 176

Offset: 0

Hugo van der Sanden, Feb 26 2005

Inspired by discussion of "True so far" from Eric Angelini (A102357).

Cf. A102823-A102830, A102357.

A102828 "True already", base 2, start 1: a(n) is the least integer such that the sequence up to a(n-1) written in base 2 contains floor(a(n)/2) copies of the digit a(n) % 2, with a(0) = 1.

Original entry on oeis.org

1, 3, 7, 13, 19, 25, 31, 41, 47, 57, 65, 69, 75, 83, 91, 101, 109, 119, 131, 137, 143, 153, 161, 167, 177, 185, 195, 203, 213, 223, 237, 249, 261, 267, 275, 283, 293, 301, 311, 323, 331, 341, 351, 365, 377, 389, 397, 407, 419, 429, 441, 453, 463, 477, 491

Offset: 0

Hugo van der Sanden, Feb 26 2005

Inspired by discussion of "True so far" from Eric Angelini (A102357).

Cf. A102823-A102830, A102357.

cp's OEIS Frontend

A102823 "True already", base 10, start 0: a(n) is the least integer such that the sequence up to a(n-1) written in base 10 contains floor(a(n)/10) copies of the digit a(n) % 10, with a(0) = 0.

Views

Author

Keywords

Comments

Examples

References

Crossrefs

A102830 "True already", base 4, start 1: a(n) is the least integer such that the sequence up to a(n-1) written in base 4 contains floor(a(n)/4) copies of the digit a(n) % 4, with a(0) = 1.

Views

Author

Keywords

References

Crossrefs

A102850 Non-monotonic "True so far" sequence: In the first n terms, the digit (a(n) mod 10) occurs floor(a(n)/10) times; a(n) is the smallest such number.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

A102824 "True already", base 2, start 0: a(n) is the least integer such that the sequence up to a(n-1) written in base 2 contains floor(a(n)/2) copies of the digit a(n) % 2, with a(0) = 0.

Views

Author

Keywords

References

Crossrefs

A102827 "True already", base 10, start 1: a(n) is the least integer such that the sequence up to a(n-1) written in base 10 contains floor(a(n)/10) copies of the digit a(n) % 10, with a(0) = 1.

Views

Author

Keywords

Comments

Examples

References

Crossrefs

Programs

Maple

A102829 "True already", base 3, start 1: a(n) is the least integer such that the sequence up to a(n-1) written in base 3 contains floor(a(n)/3) copies of the digit a(n) % 3, with a(0) = 1.

Views

Author

Keywords

References

Crossrefs

A102815 "False so far" sequence.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

PARI

Formula

A102825 "True already", base 3, start 0: a(n) is the least integer such that the sequence up to a(n-1) written in base 3 contains floor(a(n)/3) copies of the digit a(n) % 3, with a(0) = 0.

Views

Author

Keywords

References

Crossrefs

A102826 "True already", base 4, start 0: a(n) is the least integer such that the sequence up to a(n-1) written in base 4 contains floor(a(n)/4) copies of the digit a(n) % 4, with a(0) = 0.

Views

Author

Keywords

References

Crossrefs

A102828 "True already", base 2, start 1: a(n) is the least integer such that the sequence up to a(n-1) written in base 2 contains floor(a(n)/2) copies of the digit a(n) % 2, with a(0) = 1.

Views

Author

Keywords

References

Crossrefs